Convergence in $L^p$ with given properties Let $p\in [1,\infty[$, $(\Omega, \mathcal A, \mu)$ be a probability space and $(f_n)_{n\in \mathbb N}$ be a sequence of measurable functions $f_n : \Omega  \to \mathbb R$ and $f:\Omega \to \mathbb R$ with the following properties:  
$\forall \omega \in \Omega: f_n(\omega) \to f(\omega)$ and
$\sup_{n\in \mathbb N}\left \lVert f_n 1_{\{\lvert f_n \rvert > m\}}\right \rVert_p \to 0$ as $m\to \infty.$
Show that $$\lVert f_n - f \rVert_p \to 0.$$
I was able to prove one inequality:
We have $$0 = \int \lvert f-f|^p d\mu = \int \liminf_{n\to \infty} |f_n - f|^p d\mu \leq \liminf_{n\to \infty}\int \lvert f_n - f \rvert ^p d\mu = \liminf_{n\to \infty} \lVert f_n -f\rVert_p^p$$
by Fatous lemma. But how to prove the other direction? The second property is hard for me to understand. I tried to get a bound involving the measure of a set since $\mu $ is finite but feel like I need a hint.
 A: I'll take $p=1$ for simplicity. Claim: $\{f_n\}$ is uniformly integrable.
Proof: Let $\epsilon>0.$ Choose $m$ such that $\int_{|f_n|>m} |f_n| <\epsilon/2$ for all $n.$ Set $\delta = \epsilon/(2m).$ If $\mu(E) < \delta,$ then
$$\int_E |f_n| = \int_{E\cap \{|f_n|\le m\}} |f_n| + \int_{E\cap \{|f_n|> m\}} |f_n| \le m\cdot \delta + \epsilon/2 <\epsilon.$$
This proves the claim. Fatou's lemma then makes it clear that $\{f_n\}\cup \{f\}$ is uniformly integrable. Hence $\{ |f_n-f|\}$ is uniformly integrable.
Let $\epsilon>0$ again. Choose $\delta >0$ such that $\mu(E) < \delta$ implies $\int_E |f_n-f| < \epsilon$ for all $n.$ By Egorov, there exists a set $E, \mu(E) < \delta,$ such that $f_n\to f$ uniformly on $\Omega \setminus E.$ Thus
$$\int_{\Omega} |f_n-f| \le \int_{\Omega \setminus E} |f_n-f| + \int_E |f_n-f| < \epsilon.$$
The first integral on the right $\to 0$ by uniform convergence and the fact that we're on a finite measure space. Thus $\limsup \int_{\Omega} |f_n-f| \le \epsilon.$ Since $\epsilon$ was arbitrary, we have $\int_{\Omega} |f_n-f| \to 0$ as desired.
A: Suppose $M_0 \ge 0$ such that $\sup_{n\in \mathbb N}\left \lVert f_n 1_{\{\lvert f_n \rvert > M_0\}}\right \rVert_p \le 1$.


*

*$(||f_n||_p)_n$ is uniformly bounded.
$$\sup_{n\in \mathbb N}\left \lVert f_n \right \rVert_p \le \sup_{n\in \mathbb N}\left \lVert f_n 1_{\{\lvert f_n \rvert \le M\}}\right \rVert_p + \sup_{n\in \mathbb N} \left \lVert f_n 1_{\{\lvert f_n \rvert > M\}}\right \rVert_p \le M_0 + 1$$

*$f \in L^p(\Omega, \mathcal A, \mu)$: due to pointwise convergence $f_n \to f$, $|f| = \lim\limits_{n\to\infty}|f_n|$, so $|f| = \limsup_n |f| \le \sup_n |f_n|$.
$$||f||_p \le \sup ||f_n||_p \le M_0 + 1$$

*$f_n \to f$ in $L^p(\Omega, \mathcal A, \mu)$:  For any $M>0$,
\begin{align}
& ||f_n - f||_p \\
&\le ||(f_n - f) 1_{\{|f_n-f| \le \epsilon\}}||_p + ||(f_n - f) 1_{\{\epsilon < |f_n-f| \le 2M\}}||_p + ||(f_n - f) 1_{\{|f_n-f| > 2M\}}||_p \\
&\le \epsilon + 2M \mu \{|f_n-f|>\epsilon\} + || (|f_n| + |f|) 1_{\{|f_n| \vee |f| > M\}} ||_p \\
&\le \epsilon + 2M \mu \{|f_n-f|>\epsilon\}
+ 2 \underbrace{|||f_n| 1_{\{|f_n|>M\}}||_p}_{|f_n| \text{ is larger}}
+ 2 \underbrace{|||f| 1_{\{|f|>M\}}||_p}_{|f| \text{ is larger}} \\
&\le \epsilon
+ 2 \underbrace{M\mu \{|f_n-f|>\epsilon\}}_{\substack{\text{convergence a.e implies} \\ \text{convergence in measure} \\ \text{in probability space}}} +
2 \underbrace{\sup_{n\in \mathbb N}\left \lVert f_n 1_{\{\lvert f_n \rvert > M \}}\right \rVert_p }_{\substack{\text{given: no escape from} \\ \text{vertical infinity}}}
+ 2 \underbrace{|||f| 1_{\{|f|>M\}}||_p}_{f \in L^p(\Omega,\cal A, \mu)}
\end{align}
Let $M > 0$ be sufficiently large so that


*

*$\sup_{n\in \mathbb N}\left \lVert |f_n| 1_{\{\lvert f_n \rvert > M \}}\right \rVert_p < \epsilon$

*$|||f| 1_{\{|f|>M\}}||_p < \epsilon$
Let $N \in \Bbb{N}$ be sufficiently large so that for all $n \ge N$, $\mu\{|f_n-f|>\epsilon\} \le \dfrac{\epsilon}{M}$
Hence, $||f_n - f||_p \le 7 \epsilon$ for all $n \ge N$.  This shows $f_n \xrightarrow[n\to\infty]{L^p(\Omega,\cal A, \mu)} f$
Remarks:


*

*Terence Tao describes the second property in the question body
as "no escape to vertical infinity".

*I've skipped one step to simplify writings of indicator functions.  Please refer to my discussions with saz in the comments below.

*We have actually proven the result with a weaker assumpion: convergence in measure $f_n \stackrel{\mu}{\to} f$ instead of convergence almost everywhere, based on the assumption that $\mu(\Omega) < +\infty$.


*

*The advantage of this approach is that we don't need almost everywhere convergence, which is necessary in order to apply Egorov's theorem.

*Dropping the condition $\mu(\Omega) < +\infty$ invalidates this answer as the RHS of
$$||(f_n - f) 1_{\{|f_n-f| \le \epsilon\}}||_p \le \epsilon \mu(\Omega)$$
is no longer finite.


